Or is it always assumed that an observer in an inertial frame of reference thinks that he or she is at rest, and that it is only others - observers and frames of reference - that are moving? In examples which depict a moving ball and a traveling photon in a car of a train, which is itself moving at constant velocity along a train track, can an observer in the train car know that he or she is moving?
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2If the observer has studied he or she knows that velocity is relative and thus "at rest"ness is also relative. – dmckee --- ex-moderator kitten Mar 19 '17 at 05:48
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If observer A thinks he is at rest and observer B is moving, he is right. Observer B, who thinks he is at rest and A is moving, is also right.
You can choose any inertial frame of reference. When you do, motion as viewed by that frame is correct.
This extends past velocity. Observer A says he is not moving. He is in a particular place at time t1 and the same place at t2. Observer B has a different view. A is moving. He is in two different places at those two times. Both A and B are right.
Update - Some more explanation is in order.
Suppose space was completely empty. You could not tell one point from another. They are all completely alike. Likewise, you could not tell one direction from another. All times would be alike.
Suppose you (observer A) were the only object in space. Now you can tell one point from all the others. That point is the one where you are. And that is the only way to tell points apart.
You can tell that you are not rotating if there are no centrifugal forces pulling one part of you away from another part.
If you have rods, you can set up a coordinate system. You can now set up the rods to identify any point and give it a name based on how far it is from you. You can call the point where you are at a given time the origin or $(0,0,0)$.
As long as you and your rods are rigid, there is very little point to talking about time. The universe is the same at all times.
But you cannot say if you are moving or not. All points are alike. You cannot say whether or not the origin is at the same point in space at two different times. At any given time, you can say the distance between the points named $(0,0,0)$ and $(1,0,0)$ is $1$.
Things change if the universe contains observers A and B. Both A and B can set up their own coordinate systems and give their own names to points. Neither A nor B know if their origins occupy the same point in space at two times. But they can measure the distance between the two origins, and see that it changes.
You can say that B's origin is at a point named $(x_1, y_1, z_1)$ at time $t_1$, and $(x_2, y_2, z_2)$ at time $t_2$. You can say that B is moving with respect to you. That is different from saying B is moving.
B is completely free to you his own coordinate system - his own names for points - and conclude that you are moving with respect to him.
mmesser314
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Both A and B are at rest, They move relatively to each other, and they are at rest to each other. – Mar 19 '17 at 09:01
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@mmesser314: What about those diagrams, such as Minkowski diagrams, which seem to take a global view of all frames and observers, outside all the frames depicted within the diagram itself? In such diagrams, usually one such frame is asserted to be "at rest"; the other is "moving". Then it is claimed that the situation is reversible: each observer could just as well be moving as being at rest. Then the creator and the reader of the diagram are assuming a "privileged" viewpoint, which would entail that both A and B are wrong, since neither is in the at-rest state he thinks he alone is in. – Sumwun Yumaynotno Mar 19 '17 at 23:46
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@SumwunYumaynotno - It is not so much a privileged viewpoint as a chosen viewpoint. A can create a diagram that shows his timeline as vertical and x axis as horizontal. A is depicting his frame as the one at rest. He can draw B's frame, which is moving with respect to the A frame. But B is not wrong to regard his frame as fixed and A as moving. B can create his own diagram which represents the world from the B frame. Both diagrams can be used to do correct physics. – mmesser314 Mar 20 '17 at 00:43
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@mmesser314 It is a privileged viewpoint because it is not subject to the limitations that observers within the diagram are subject to: it effectively eliminates the passage of time by allowing one to depict past, present, and future all at once; it can “know” what is going on in space-time without having to wait for any light signal to reach one’s eyes; it “knows” what any depicted observer “must see” and “must conclude” before that observers moves along his world-line. No real-life observer has this ability. – Sumwun Yumaynotno Mar 20 '17 at 01:12
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If – and only if – an actual observer already knows about the theory of relativity and all it implies, can that observer assume the outsider’s point of view. It is the difference between perception and conception. – Sumwun Yumaynotno Mar 20 '17 at 01:13
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@mmesser314 Sorry, but I find this to be yet another implausible, inconsistent thought-experiment that I keep running into. You say “Now you can tell one point from all the others.” A few sentences later, “All points are alike. You cannot say whether or not the origin is at the same point in space at two different times.” This is a flat self-contradiction. You can’t rely on “centrifugal force” because there literally is no such thing; it is acknowledged to be a just convenient fiction. You’re miles away from the question I asked: There is a railroad track, there is a moving train. – Sumwun Yumaynotno Mar 20 '17 at 04:27
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There are a moving ball and a moving photon in a car in the train; you can call them frames themselves, if you want. The train itself could be moving along inside a car on yet another train, and so forth. They could be be moving in the same direction or in opposite directions. You can perform various calculations to determine relative velocities. But I want to know: who can do such calculations? If you’re residing inside a frame, in a state of inertial motion (not thrashing around at the beginning of the universe, as your thought-experiment lays it out) then can you know you’re moving? – Sumwun Yumaynotno Mar 20 '17 at 04:28
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If you don’t, then, it seems to me, only a hypothetical observer with an omniscient point of view can know about such things – the big, all-inclusive space-time picture. If you can know, then you are assuming the point of view of such an observer; you are imagining, conceiving, what you cannot perceive. – Sumwun Yumaynotno Mar 20 '17 at 04:29
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He cannot know, but he can ascribe himself either a state of proper rest or state of proper motion. The reflections below are to demonstrate, that two observers cannot ascribe themselves the state of rest simultaneously and that observations by moving observes are different from that who is at rest.
If two observers move relatively to each other with relative velocity v = 0.9 c ,in no way they can be at rest simultaneously, since they move relatively to each other. Whatever each of them thinks.
Relativistic observers A and B are equivalent in that sense, that each of them can think that he is „at rest“ and conduct observations from his rest frame, filled by Einstein – synchronized clocks. But A is at rest, B should not change his opinion of his state of motion, and consider himself being in motion.
Thus, if observer A ascribes himself state of proper rest, he adds another clock A2 at certain distance from his own A1, synchronizes clocks by light, admitting that velocity of light is c (Einstein synchronization convention).
In this rest frame velocity of light is considered to be c in all directions.
Then observer A measures dilation of observer's B clock. Clock B1 passes by clock A1 first and clock A2 then. Two spatially separated clocks A1 and A2 will measure longer period of time than clock B1. If clocks B1 and A1 showed 12 PM at the meeting both, A2 will show 6 PM and B1 3 PM when their meet.
But, let‘s observer B ascribes himself state of proper motion. What he does? He does not introduce his own rest frame (i.e. does not place another clock at certain distance from his own) but uses A‘s rest frame. He notices, that according to his clock, time in reference frame A runs gamma times faster than his own.
Reciprocity of observations and those miraculous effect like “You shorter than I and I shorter than you“ appear as soon as every relativistic observer covers space with his own rest frame with synchronizes clocks by the same method (Einstein“s synchronization). If they choose one frame and stay within that arbitrary chosen frame, moving observer would measure that time in reference frame runs faster and measuring rod „at rest“ stretches.
Now imagine that observers A and B play some kind of ping pong with photon. They agree, that they will launch a photon transversely to direction of motion. Each of them possesses a tube. This tube combines a laser pointer and telescope. Observer B ascribes himself state of proper rest and stays in the origin and launches a photon straight up along y axis.
Observer A moves at parallel line to axis x. Photon had to be launched a bit prior than observer A crosses point Y of y axis, because it takes the same time for photon and observer A to meet at point Y.
Can observer A ascribe himself state of proper rest too? No, he can‘t. Why? Because he will not see the photon then.
Observer A MUST tilt his telescope at oblique angle $\theta_A$ if he wants to see the light. It is like Bradley tilted his telescope a bit into front and observed, that distant stars change their position. It is because of aberration of light. This angle depends on relative velocity of observer, and source of light appears in the front of him. Angle of emission and angle of reception are tied with relativistic aberration formula.
$$ \cos {\theta_A} = \frac {\cos {\theta_B} - \frac v c} {1- \frac v c \cos \theta_B} $$
Let‘s consider, that observer A rotates around observer B. Observer B emits light. In this case observer A always has to keep his telescope at oblique angle, because flux of radiation comes at oblique angle to A. Photons will always be blueshifted at $\gamma$.
If A emits light, the light will always come at right angle to B. Photons will always be redshifted at $\gamma$.
This article in Wikipedia (chapter Transverse Doppler Effect) explains, that moving observer (light received at points of closest approach) will see blueshift (clock runs faster than your own) of radiation, and observer „at rest“ (light emitted at points of closest approach) will see redshift (clock runs slower than your own) respectively.
That is fully in accordance with dilation of that clock, which is "in motion" in chosen reference frame.
https://en.wikipedia.org/wiki/Relativistic_Doppler_effect
You may ask, what will happen if the both observers will chose a reference frame, in which the both move with equal velocities. Well, they will not see any time dilation, because A has to tilt his laser pointer backward and B has to tilt his telescope forward at equal angles.That simply means, that their clock dilate at the same magnitude since they move with equal velocities relatively to one mutual REFERENCE FRAME.
http://mathpages.com/home/kmath587/kmath587.htm
"A simple way of expressing this is to point out that the null frequency shift occurs for the pulse that travels the shortest distance from emitter to receiver, and this pulse is obviously neither emitted nor received at the point when the emitter and receiver are at their point of closest approach. Since the pulse requires an amount of time to traverse the distance between emitter and receiver, it must be emitted slightly before the point of closest approach and received slightly after. Of course, the spatial distance traveled by this pulse (for emitter and receiver on parallel paths) is equal to the minimum distance between the emitter and receiver, i.e., the perpendicular distance between the paths"
Animation:
https://www.youtube.com/watch?v=hnphFr2Iai4
https://www.youtube.com/watch?v=AGMINcBYojc
Observers A and B can choose a frame, in which A is at rest. A keeps tube straight up, B keeps tube „into front“. A measures dilation of B‘s clock, B measures that A‘s clock run faster.
Observers can choose a frame, in which B is at rest. B keeps tube straight up, A keeps tube „into front“. B measures that A‘s clock run slower, A measures that B‘s clock runs faster.
Observers can choose a frame, in which A and B move with equal velocities. A keeps tube backward, B keeps tube into front. They measure no dilation
They cannot keep their tubes „up“ and „up“ respectively. They cannot be at rest together, since they relatively move. Frame in which they both are at rest does not exist.
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When you assert “in no way they can be at rest simultaneously, since they move relatively to each other”, you are taking a privileged, outside view, which “knows” what the situation “really” is, regardless of “whatever each of them thinks”. When you say “Two spatially separated clocks A1 and A2 will measure longer period of time than clock B1.”, I have to ask: exactly who is doing this comparison? A? B? Or is it the omniscient outside observer who knows the objective reality that neither A nor B can know? – Sumwun Yumaynotno Mar 20 '17 at 00:30
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If B “ascribes himself state of proper motion”, then he is no longer relying upon his own observations and measurements; he is deliberately circumventing them. I don’t understand how B could “use A’s rest frame”. This would mean that he already knows about the relativity of simultaneity, time dilation, and so forth. I don’t think that any Minkowski diagram is meant to impute such theoretical knowledge to any observer depicted in the diagram. I don’t agree that “You shorter than I and I shorter than you“ appears at all; it is a conclusion reached by an omniscient outside observer. – Sumwun Yumaynotno Mar 20 '17 at 00:32
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I don’t see that any actual observer has the ability to “choose” a frame of reference; one is stuck in a frame of reference that governs what he sees. If one could at will choose a frame, one would no longer be an observer, but rather a theorist. I believe that you are mixing up perception with conception. I can’t follow (or don’t feel like trying to follow) the rest of your elaborate thought experiments. – Sumwun Yumaynotno Mar 20 '17 at 00:34
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"exactly WHO is doing this comparison?" It seems we have to start with basics of SR. Observer in SR is not a physical person. It is the whole rest frame. Yes, to some extent omniscient observer. Reference frame is a at least two observers at certain distance from each other, each possesses a clock, and these clocks synchronized by Einstein convention. Thus, observer "creates" his own rest frame. https://en.wikipedia.org/wiki/Observer_(special_relativity). http://www.pstcc.edu/departments/natural_behavioral_sciences/Web%20Physics/Chapter039.htm – Mar 20 '17 at 07:14
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I explain further, that, in my opinion, two relativistic observers are tied. I avoid procedure of creating rest frames and conducting measurements by synchronized clocks, because these procedures based on convention. They can synchronize clock as they wish, right it or wrong. I look at the problem through a prism of a real physical experiment. Aberration of light does not allow to observers to think whatever they want and to behave as they please. They must to come to agreement, who moves and who is at rest, if they want to see a beam of laser light. – Mar 20 '17 at 07:24
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Maybe this post will help http://physics.stackexchange.com/questions/318825/how-are-coordinate-systems-built-physically – Mar 20 '17 at 07:29